对于此类问题，如何使用左连接编写 mySQL？

发布于 2024-10-21 02:21:50 字数 346 浏览 4 评论 0原文

如何使用左连接构建 mySQL 查询来解决此问题？

Feed Table
-id
-lastpost_id (The id of the user from USER TABLE who last posted)
-firstpost_id (The id of the user from USER TABLE who first posted)

User Table
-id
-name

我希望得到一张这样的桌子

Result 
-id
-lastpost_id
-lastpost_name
-firstpost_id
-firstpost_name

原文

How do I construct the mySQL query using left-join for this problem?

Feed Table
-id
-lastpost_id (The id of the user from USER TABLE who last posted)
-firstpost_id (The id of the user from USER TABLE who first posted)

User Table
-id
-name

I wish to get a table like that

Result 
-id
-lastpost_id
-lastpost_name
-firstpost_id
-firstpost_name

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口干舌燥 2024-10-28 02:21:50

select *, a.name as lastpost_name, b.name as firstpost_name from Feed f 
      left join user a on f.lastpost_id=a.id 
      left join user b on f.firstpost_id=b.id

技巧是左连接，然后将表别名为 a 或 b。这样您就可以对同一个表进行多个联接

select *, a.name as lastpost_name, b.name as firstpost_name from Feed f 
      left join user a on f.lastpost_id=a.id 
      left join user b on f.firstpost_id=b.id

The trick is to left join and then alias the table as a or b. That way you can make multiple joins to the same table

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